Question

Question: Earth’s mass is estimated to be $\mathbf{5}\mathbf{.}\mathbf{98}\mathbf{\times }{\mathbf{10}}^{\mathbf{24}}\mathbf{}\mathbf{kg}$, and titanium represents $\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{\%}$by mass of this total. (a) How many moles of are present? (b) If half of the $\mathbf{Ti}$is found as ilmenite $\mathbf{\left(}{\mathbf{FeTiO}}_{\mathbf{3}}\mathbf{\right)}$, what mass of ilmenite is present? (c) If the airline and auto industries use $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}$tons of $\mathbf{Ti}$per year, how many years would it take to use up all the$\mathbf{Ti}\mathbf{}\left(1\mathrm{ton}\times 2000\mathrm{lb}\right)$?

Short answer

1. $6.2465x{10}^{22}$moles of Titanium are present in the Earth.
2. The total amount of ilmenite present is $4.7383\times {10}^{24}\mathrm{g}$.
3. To consume all of the Titanium on Earth, $3.432x{10}^{13}$Years are required.

Step-by-step solution

Explanation of the concept

The study of carbon compounds is known as organic chemistry. • Organic Molecules and the Origin of Life on Earth o Stanley Miller: concluded in his experiment that complex organic molecules could arise spontaneously under conditions thought to exist on early Earth at the time. The experiment also supported that idea that abiotic synthesis of organic compounds could have been an early stage in the origin of life. Carbon atoms can form diverse molecules by bonding to 4 other atoms. • Electron arrangement.

Calculating the total mass of the earth

$47.8670\mathrm{g}/\mathrm{mol}$Given the total mass of the Earth and the mass percentage Titanium is the mass of the$\mathrm{Ti}$initially calculated:

$$\begin{gathered} \mathrm{m}\left(\mathrm{Ti}\right)=\frac{0.05\%}{100\%}\times 5.98\times {10}^{24}\mathrm{kg}\times \frac{{10}^3\mathrm{g}}{1\mathrm{kg}} \\ \mathrm{m}\left(\mathrm{Ti}\right)=2.99\times {10}^{24}\mathrm{g} \end{gathered}$$

Because the molar mass of Ti is$47.8670\mathrm{g}/\mathrm{mol}$the quantity Ti present

$$\begin{gathered} \mathrm{n}\left(\mathrm{Ti}\right)=\frac{\mathrm{m}}{{\mathrm{M}}_{\mathrm{R}}} \\ \mathrm{n}\left(\mathrm{Ti}\right)=\frac{2.99\times {10}^{24}\mathrm{g}}{47.8670\mathrm{g}/\mathrm{mol}} \\ \mathrm{n}\left(\mathrm{Ti}\right)=6.2465\times {10}^{22}\mathrm{mol} \end{gathered}$$

Therefore,$6.2465x{10}^{22}$moles of Ti are present on Earth.

Calculating the total amount of ilmenite present

If half of Ti is discovered as ilmenite $\left({\mathrm{FeTiO}}_3\right)$and ilmenite only has Ti. The amount of ilmenite present in the molecule is per atom.

$$\begin{gathered} n\left(FeTi{O}_3\right)=\frac{1}{2}n\left(Ti\right) \\ n\left(FeTi{O}_3\right)=\frac{1}{2}\times 6.2465\times {10}^{22}mol \\ n\left(FeTi{O}_3\right)=3.1235\times {10}^{22}mol \end{gathered}$$

Ilmenite has a molar mass of $151.71\mathrm{g}/\mathrm{mol}$. The total amount of ilmenite present is

$$\begin{gathered} \mathrm{m}\left({\mathrm{FeTiO}}_3\right)=\mathrm{n}\left({\mathrm{FeTiO}}_3\right)\times {\mathrm{M}}_{\mathrm{R}}\left({\mathrm{FeTiO}}_3\right) \\ \mathrm{m}\left({\mathrm{FeTiO}}_3\right)=3.123\times {10}^{22}\mathrm{mol}\times 151.71\mathrm{g}/\mathrm{mol} \\ \mathrm{m}\left({\mathrm{FeTiO}}_3\right)=4.738\times {10}^{24}\mathrm{g} \end{gathered}$$

Therefore, the total amount of ilmenite present is $4.783\times {10}^{24}\mathrm{g}$.

Converting the Consumption rate

To begin, convert the consumption rate to SI units:

$$\begin{gathered} 1\mathrm{ton}=2000\mathrm{lbs} \\ 1\mathrm{lbs}=453.592\mathrm{g} \end{gathered}$$

As a result, the total annual Ti consumption is

$$\begin{gathered} \mathrm{m}\left(\mathrm{consumed}\right)=1.00\times {10}^5\mathrm{ton}\times \frac{2000\mathrm{lbs}}{1\mathrm{ton}}\times \frac{453.592\mathrm{g}}{1\mathrm{lbs}} \\ \mathrm{m}\left(\mathrm{consumed}\right)=8.71184\times {10}^{10}\mathrm{g} \end{gathered}$$

Thus, given the total mass of titanium present and the annual consumption, the time required to consume all of the Ti is:

localid="1663324793755" $$\begin{gathered} \mathrm{Time}=\frac{2.99\times {10}^{24}\mathrm{g}}{8.71184\times {10}^{10}\mathrm{g}/\mathrm{year}} \\ \mathrm{Time}=3.432\times {10}^{13}\mathrm{years} \end{gathered}$$

To consume all of the Ti on Earth, $3.432x{10}^{13}$Years are required.